Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 1;
struct edge {
  int to, nxt;
} e[N << 1];
int head[N];
int edge_cnt;
void add_edge(int u, int v) {
  e[++edge_cnt] = (edge){v, head[u]};
  head[u] = edge_cnt;
}
const int LOG = 17;
int n, m, q;
int bz[LOG + 1][N][11];
vector<int> tag[N];
int father[LOG + 1][N];
int dep[N];
void Dfs(int u, int Father) {
  father[0][u] = Father;
  dep[u] = dep[Father] + 1;
  for (int i = 1; i <= LOG; ++i) father[i][u] = father[i - 1][father[i - 1][u]];
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == Father) continue;
    Dfs(v, u);
  }
}
int lca(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  for (int i = LOG; i >= 0; --i)
    if (dep[father[i][u]] >= dep[v]) u = father[i][u];
  if (u == v) return u;
  for (int i = LOG; i >= 0; --i)
    if (father[i][u] != father[i][v]) u = father[i][u], v = father[i][v];
  return father[0][u];
}
int c[11];
void merge(int* a, int* b) {
  int x = 1, y = 1;
  c[0] = 0;
  for (int i = 1; i <= 10; ++i) {
    if (x == a[0] + 1 && y == b[0] + 1)
      break;
    else if (x == a[0] + 1)
      c[++c[0]] = b[y++];
    else if (y == b[0] + 1)
      c[++c[0]] = a[x++];
    else
      c[++c[0]] = a[x] < b[y] ? a[x++] : b[y++];
  }
  for (int i = 0; i <= c[0]; ++i) a[i] = c[i];
}
void merge(int* a, vector<int> b) {
  int x = 1, y = 0;
  c[0] = 0;
  for (int i = 1; i <= 10; ++i) {
    if (x == a[0] + 1 && y >= (int)b.size())
      break;
    else if (x == a[0] + 1)
      c[++c[0]] = b[y++];
    else if (y >= (int)b.size())
      c[++c[0]] = a[x++];
    else
      c[++c[0]] = a[x] < b[y] ? a[x++] : b[y++];
  }
  for (int i = 0; i <= c[0]; ++i) a[i] = c[i];
}
void Dfs_1(int u) {
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == father[0][u]) continue;
    Dfs_1(v);
    merge(bz[0][v], tag[u]);
  }
}
void climb_up(int u, int v, int* ans) {
  for (int i = LOG; i >= 0; --i) {
    if (dep[father[i][u]] > dep[v]) {
      merge(ans, bz[i][u]);
      u = father[i][u];
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> q;
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    add_edge(u, v);
    add_edge(v, u);
  }
  Dfs(1, 0);
  for (int i = 1; i <= n; ++i) tag[i].clear();
  for (int i = 1; i <= m; ++i) {
    int x;
    cin >> x;
    tag[x].push_back(i);
  }
  Dfs_1(1);
  for (int i = 1; i <= n; ++i) {
    sort(bz[0][i] + 1, bz[0][i] + bz[0][i][0] + 1);
    sort(tag[i].begin(), tag[i].end());
    tag[i].resize(min((int)tag[i].size(), 10));
  }
  for (int i = 1; i <= LOG; ++i)
    for (int j = 1; j <= n; ++j) {
      for (int k = 0; k <= bz[i][j][0]; ++k) bz[i][j][k] = bz[i - 1][j][k];
      merge(bz[i][j], bz[i - 1][father[i - 1][j]]);
    }
  int ans[11];
  for (int qt = 1; qt <= q; ++qt) {
    int u, v, k;
    cin >> u >> v >> k;
    int _lca = lca(u, v);
    ans[0] = 0;
    merge(ans, tag[u]);
    if (u != v) merge(ans, tag[v]);
    if (u == _lca) {
      climb_up(v, u, ans);
    } else if (v == _lca) {
      climb_up(u, v, ans);
    } else {
      merge(ans, tag[_lca]);
      climb_up(u, _lca, ans);
      climb_up(v, _lca, ans);
    }
    cout << min(k, ans[0]) << ' ';
    for (int i = 1; i <= min(k, ans[0]); ++i) cout << ans[i] << ' ';
    cout << '\n';
  }
  return 0;
}